%!TEX program = xelatex
\documentclass[t,12pt,aspectratio=169]{beamer} % 16:9 宽屏比例，适合现代投影
%\usepackage{ctex} % 中文支持
\usepackage{amsmath, amsthm, amssymb, bm} % 数学公式与符号
\usepackage{graphicx}
\usepackage{hyperref}
\usepackage{color}

% 设置段落间距
\usepackage{setspace}
\onehalfspacing
\setlength{\parskip}{1em}  % 增加段落之间的间距为1em

% 每页增加与上面标题行的距离
\addtobeamertemplate{frametitle}{}{\vspace*{0.7em}}

% 主题设置（推荐简洁风格）
\usetheme{Madrid}
\usecolortheme{default} % 可选：seahorse, beaver, dolphin 等

%\linespread{1.3}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 信息设置
\title{Chapter 15: Embeddings}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

\begin{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 封面页
\begin{frame}
  \titlepage
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 目录页
\begin{frame}{Contents}
  \tableofcontents
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 0
%\section{INTRO.}
\begin{frame}{intro. }

In the previous chapter we defined the inverse image of a module and calculated a formula for its dimension and multiplicity under projections. 

In this chapter we turn to embeddings. 

As we shall see, the behaviour of the inverse image under embeddings is a lot less regular than under projections.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 1
\section{The Standard Embedding}
\begin{frame}[allowframebreaks]{A. }

In this chapter we shall retain the notation of 14.1.2. 

We begin by considering the embedding $\iota: X \to X \times Y$ defined by $\iota(X) = (X,0)$, where $0$ denotes the origin of $Y$. 

We will call $\iota$ the standard embedding. 

The comorphism $\iota^{\sharp}: K[X,Y] \to K[X]$ is defined by $\iota^{\sharp}(g(X,Y)) = g(X,0)$. 

It may be used to make $K[X]$ into a $K[X,Y]$-module; as such, $K[X]$ is generated by $1$, since
\[ g(X,Y) \cdot 1 = \iota^{\sharp}(g(X,Y)) \cdot 1 = g(X,0). 
\]
The annihilator of $1$ in $K[X,Y]$ is the ideal generated by $y_1,\ldots,y_m$. 

Denoting this ideal by $(Y)$ we have that $K[X] \cong K[X,Y]/(Y)$ as $K[X,Y]$-modules.


Now consider a left $A_{m+n}$-module $M$. 

Since $A_n \subseteq A_{m+n}$, the $A_{m+n}$-module $M$ is also an $A_n$-module. 

The elements of $A_n$ commute with the variables $y_1,\ldots,y_m$, thus $(Y)M$ is an $A_n$-submodule of $M$. 

Hence $M/(Y)M$ is an $A_n$-module. 

Note however that it is not an $A_{m+n}$-submodule. 

We want to show that
\[ \iota^{*} M = K[X] \otimes_{K[X,Y]} M \cong M/(Y)M \]
as $A_n$-modules.


If $u \in M$, let $\overline{u}$ denote its image in $M/(Y)M$. 

Define the map
\[ \phi: \iota^{*} M \to M/(Y)M \]
by $\phi(q \otimes u) = q \overline{u}$. 

It is a homomorphism of $K[X]$-modules. 

We want to show that the action of the derivations $\partial_{x_1},\ldots,\partial_{x_n}$ is compatible with $\phi$. 

We have that
\[ \partial_{x_i}(q \otimes u) = \frac{\partial q}{\partial x_i} \otimes u + \sum_{k=1}^{n} q \frac{\partial x_k}{\partial x_i} \otimes \partial_{x_k} u, \]
and so
\[
\partial_{x_i}(q \otimes u) = \frac{\partial q}{\partial x_i} \otimes u + q \otimes \partial_{x_i}u.
\]

The right hand side is mapped by $\phi$ onto
\[
\frac{\partial q}{\partial x_i} \overline{u} + q \overline{\partial_{x_i}u},
\]

which is equal to
\[
\partial_{x_i}(q \overline{u}) = \partial_{x_i}(\phi(q \otimes u)).
\]

Thus $\phi$ is an isomorphism of $A_n$-modules.

Summing up: $\iota^*M \cong M/(Y)M$ as $A_n$-modules. 

Since $M/(Y)M$ is a quotient of $A_n$-modules, the action of $x$'s and $\partial$'s is the natural one.

Let us specialize this construction to the standard embedding $\iota : K \to K^2$, which takes $x$ to $(x,0)$. 

Let $M$ be the ring $A_2$ itself, considered as a left $A_2$-module. 

Applying the above results, we have that $\iota^*M$ is the quotient $A_2/yA_2$ taken as a left $A_1$-module. 

Note that although $M = A_2$ is a finitely generated left $A_2$-module, the module $\iota^*M$ is not finitely generated. 

In fact, $\iota^*M$ is a free $A_1$-module whose basis is the image of the powers of $\partial_y$ in $A_2/yA_2$. 

Since this basis is infinite, $\iota^*M$ is not a finitely generated left $A_1$-module.


We conclude from this example that the inverse image under embeddings does not preserve the noetherian property. 

This is very important, since we have defined dimensions only for noetherian modules. 

In particular there can be no general formula for the dimension of an inverse image under an embedding. 

Despite this, holonomic modules are preserved by inverse images under embeddings. 

This mystifying fact will have to wait until Ch. 18 for an explanation.


It will be necessary, for future purposes, to consider a more general kind of embedding, defined as a composition of the standard embedding and a polynomial isomorphism. 

Before we do this we must study the behaviour of the inverse image under composition of polynomial maps.



\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 2
\section{Composition}
\begin{frame}[allowframebreaks]{B. }

Let us first consider what happens to the change of rings construction of Ch. 14, §1 under composition. 

Let $R$, $S$ and $T$ be rings and $\phi : R \to S$ and $\psi: S \to T$ be ring homomorphisms. 

Let $M$ be a left $R$-module. 

We may turn $M$ into a $T$-module in two different ways. 

On the one hand, we may apply the change of rings construction twice to get $T \otimes_{\psi} S \otimes_{\phi} M$. 

On the other hand, the ring $T$ is a right $R$-module via the homomorphism $\psi \phi$, therefore $T \otimes_{\psi \phi} M$ is a $T$-module, where
\[
t \otimes ru = t \psi \phi(r) \otimes u
\]
for $t \in T$, $r \in R$ and $u \in M$. 

Using Propositions 4.1 and 4.4 of Ch. 12, we conclude that these $T$-modules are isomorphic; thus,
\[
T \otimes_{\psi} S \otimes_{\phi} M \cong T \otimes_{\psi \phi} M.
\]

It is not difficult to give a direct proof of this isomorphism. 

The $T$-module $T \otimes_{\psi} S \otimes_{\phi} M$ is generated by elements of the form $t \otimes s \otimes u$, where $t \in T$, $s \in S$ and $u \in M$. 

But
\[
t \otimes s \otimes u = t \psi(s) \otimes 1 \otimes u.
\]

Define a $T$-module homomorphism
\[
\theta: T \otimes_{\psi} S \otimes_{\phi} M \to T \otimes_{\psi \phi} M
\]
by $\theta(t \otimes s \otimes u) = t \psi(s) \otimes u$. 

Note that if $r \in R$ then the element
\[
t \otimes 1 \otimes ru = t \psi \phi(r) \otimes 1 \otimes u
\]
is mapped to $t \psi \phi(r) \otimes u$. 

It is easy to construct an explicit inverse for $\theta$; hence it is an isomorphism.


We will now apply this to $A_n$-modules. 

Recall that by Theorem 4.1.1, if $F: X \to Y$ and $G: Y \to Z$ are polynomial maps, then $(GF)^{\sharp} = F^{\sharp} G^{\sharp}$. 

Note the change in the order of the maps.

\textbf{Theorem.}
Let $F: X \to Y$ and $G: Y \to Z$ be polynomial maps and $M$ be an $A_r$-module. 

Then
\[
F^{*} G^{*} M \cong (GF)^{*} M
\]
as $A_n$-modules.

\textbf{Proof:} By definition,
\[
(GF)^*M = K[X] \otimes_{K[Z]} M.
\]

In this formula $K[X]$ is a right $K[Z]$-module. 

Recall that the action of $f \in K[Z]$ on $h \in K[X]$ is given by $h(f \cdot G \cdot F)$. 

On the other hand,
\[
F^*G^*M = K[X] \otimes_{K[Y]} K[Y] \otimes_{K[Z]} M.
\]

We have already seen that these two modules are isomorphic as $K[X]$-modules. 

The isomorphism $\theta$ is defined by
\(
\theta(h \otimes g \otimes u) = h(g \cdot F) \otimes u.
\)

We must check that $\theta$ is compatible with the action of the derivatives $\partial_x$. 

By definition,
\[
\partial_{x_i}(h \otimes g \otimes u) = \frac{\partial h}{\partial x_i} \otimes (g \otimes u) + h \sum_{j=1}^{m} \left( \frac{\partial F_j}{\partial x_i} \otimes \partial_{y_j}(g \otimes u) \right)
\]
where $F_1, \ldots, F_m$ are the coordinate functions of $F$. 

Replacing $\partial_{y_j}(g \otimes u)$ by its formula, the term
\[
\sum_{j=1}^{m} \left( \frac{\partial F_j}{\partial x_i} \otimes \partial_{y_j}(g \otimes u) \right)
\]
becomes
\[
\sum_{j=1}^{m} \left( \frac{\partial F_j}{\partial x_i} \otimes \frac{\partial g}{\partial y_j} \otimes u \right) + \sum_{j=1}^{m} \sum_{k=1}^{r} \left( \frac{\partial F_j}{\partial x_i} \otimes g \frac{\partial G_k}{\partial y_j} \otimes \partial_{z_k} u \right).
\]

Applying $\theta$ to this expression:
\[
\sum_{j=1}^{m} \left( \frac{\partial F_j}{\partial x_i} \left( \frac{\partial g}{\partial y_j} \cdot F \right) \otimes u \right) + \sum_{j=1}^{m} \sum_{k=1}^{r} \left( (g \cdot F) \frac{\partial F_j}{\partial x_i} \left( \frac{\partial G_k}{\partial y_j} \cdot F \right) \otimes \partial_{z_k} u \right),
\]
which, by interchanging the summations and applying the chain rule, becomes
\[
\sum_{j=1}^{m} \left( \frac{\partial F_j}{\partial x_i} \left( \frac{\partial g}{\partial y_j} \cdot F \right) \otimes u \right) + \sum_{k=1}^{r} (g \cdot F) \left( \frac{\partial (G_k \cdot F)}{\partial x_i} \otimes \partial_{z_k} u \right).
\]

Substituting the last expression in $\theta(\partial_{x_i}(h \otimes g \otimes u))$, one obtains
\[
(g \cdot F) \frac{\partial h}{\partial x_i} \otimes u + h \sum_{j=1}^{m} \left( \frac{\partial F_j}{\partial x_i} \left( \frac{\partial g}{\partial y_j} \cdot F \right) \otimes u \right) + h \sum_{k=1}^{r} (g \cdot F) \frac{\partial (G_k \cdot F)}{\partial x_i} \otimes \partial_{z_k} u.
\]

Another application of the chain rule and Leibniz's rule to the first two terms shows that this is equal to
\[
\frac{\partial h(g \cdot F)}{\partial x_i} \otimes u + h \sum_{k=1}^{r} (g \cdot F) \left( \frac{\partial (G_k \cdot F)}{\partial x_i} \otimes \partial_{z_k} u \right),
\]
and hence to $\partial_{x_i}(h(g \cdot F) \otimes u)$. 

We have thus proved that
\[
\theta(\partial_{x_i}(h \otimes g \otimes u)) = \partial_{x_i}(h(g \cdot F) \otimes u).
\]

Thus $\theta$ is an isomorphism of $A_n$-modules.

Applying this theorem to polynomial isomorphisms we get the following corollary.

\textbf{Corollary.} 
Let $F: X \to X$ be a polynomial isomorphism and $G$ its inverse. 

If $M$ is a left $A_n$-module, then $F^* G^* M \cong M$.


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 3
\section{Embeddings Revisited}
\begin{frame}[allowframebreaks]{C. }


In this section we study a more general type of embedding. 

Let $F: X \to Y$ be a polynomial map. 

Define a new polynomial map $j: X \to X \times Y$ by $j(X) = (X, F(X))$. 

This is an injective map, as one easily checks.


We shall write $j$ as a composition of two maps, as follows. 

Let $\iota: X \to X \times Y$ be the standard embedding of §1. 

Let $G: X \times Y \to X \times Y$ be the polynomial map defined by $G(X,Y) = (X, Y + F(X))$. 

Then $G$ is bijective and $j = G \cdot \iota$.


Now let $M$ be a left $A_{m+n}$-module. 

We will calculate $j^* M$. 

By Theorem 2.1, it equals $\iota^* G^* M$. 

Let us compute $G^* M$. 

First of all, by Corollary 1.3.2, there is an automorphism $\sigma$ of $A_{m+n}$ which maps $x_i$ and $\partial_{y_j}$ to themselves and satisfies
\[
\begin{aligned}
\sigma(y_j) &= y_j - F_j(X), \\ 
\sigma(\partial_{x_i}) &= \partial_{x_i} + \sum_{j=1}^{n} (\partial F_j / \partial x_i) \partial_{y_j}.
\end{aligned}
\]


Note that $\sigma$ restricts to an automorphism of $K[X,Y]$ which we will also call $\sigma$. 

This automorphism is the inverse of the comorphism $G^{\sharp}$. 

By Lemma 14.1.1,
\[
G^* M \cong M_{\sigma}
\]
as a $K[X,Y]$-module. 

Let $\psi$ stand for this isomorphism; it satisfies $\psi(h \otimes u) = \sigma(h)u$, for $h \in K[X,Y]$ and $u \in M$.

\textbf{Theorem.} 
Let $M$ be a left $A_{m+n}$-module. 

Then $G^* M \cong M_{\sigma}$ as $A_{m+n}$-modules.

\textbf{Proof:} We know that the isomorphism holds for $K[X,Y]$-modules. 

Let us investigate the behaviour of the action of $\partial_{x_i}$ under $\psi$. 

By definition, $\partial_{x_i}(h \otimes u)$ equals
\[
\partial_{x_i}(h) \otimes u + h \otimes \partial_{x_i} u + \sum_{1}^{n} h \frac{\partial (F_j)}{\partial x_i} \otimes \partial_{y_j} u.
\]

Applying $\psi$ to this formula, we get
\begin{equation}
\sigma(\partial_{x_i}(h))u + \sigma(h)\partial_{x_i}u + \sum_{1}^{n} \sigma(h \partial F_j / \partial x_i) \partial_{y_j} u. \tag{3.2}
\end{equation}

Since $\sigma$ leaves the elements of $K[X]$ unchanged, we have that $\sigma(F_j) = F_j$. 

Thus (3.2) is equal to
\[
\sigma(\partial_{x_i}(h))u + \sigma(h)\sigma(\partial_{x_i})u.
\]

But $[\partial_{x_i}, h] = \partial_{x_i}(h)$. 

Hence
\[
\sigma(\partial_{x_i}(h)) + \sigma(h)\sigma(\partial_{x_i}) = \sigma(\partial_{x_i})\sigma(h),
\]
from which we deduce that
\[
\psi(\partial_{x_i}(h \otimes u)) = \sigma(\partial_{x_i})\sigma(h)u.
\]

Thus
\[
\psi(\partial_{x_i}(h \otimes u)) = \sigma(\partial_{x_i})\psi(h \otimes u),
\]
as required. 

One may similarly check that the action of $\partial_{y_j}$ is compatible with $\psi$.

The next corollary is a combination of Theorem 3.1 and Corollary 9.2.4.


\textbf{Corollary.} 
Let $M$ be a finitely generated left $A_{m+n}$-module. 

Then $G^* M$ and $M$ have the same dimension.

We will put all this together in a theorem.

\textbf{Theorem.} 
Let $M$ be a left $A_{m+n}$-module. 

Then
\[
j^* M \cong M_{\sigma} / (Y) M_{\sigma}
\]
as $A_n$-modules.

\textbf{Proof:} As we have seen, $j^* M \cong \iota^* G^* M$. 

By Theorem 3.1, $G^* M \cong M_{\sigma}$. 

By §1,
\[
\iota^* M_{\sigma} \cong M_{\sigma} / (Y) M_{\sigma},
\]
as claimed.



\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 4
\section{Exercises}
\begin{frame}[allowframebreaks]{D. }

\textbf{Exercise 4.1.} 
Let $\iota: K \to K^2$ be the standard embedding. 

Compute the inverse image under $\iota$ of the following $A_2$-modules.
\begin{enumerate}
    \item $A_2 / A_2 \partial_2$
    \item $K[x_1, x_2]$
    \item $A_2 / A_2 x_2$
    \item $A_2 / A_2 x_2 \partial_2$
    \item $A_2 / A_2 \partial_2^2$
\end{enumerate}

\newpage

\textbf{Exercise 4.2}

Which of the inverse images of Exercise 4.1 are finitely generated over $A_1$?

\newpage

\textbf{Exercise 4.3}

Let $\iota: X \to X \times K$ be the standard embedding, and denote by $y$ the coordinate of $K$. 

Let $A_{n+1}$ be the Weyl algebra generated by $x$'s and $\partial_x$'s and by $y$ and $\partial_y$. 

Suppose that the term of highest order of $d \in A_{n+1}$ is $\partial_y^k$. 

Show that $\iota^*(A_{n+1} / A_{n+1} d)$ is a free module of rank $k$ over $A_n$.

\newpage

\textbf{Exercise 4.4}

Keep the notations of Exercise 4.3. 

Suppose that $\mathcal{I}$ is a left ideal of $A_{n+1}$ which properly contains $A_{n+1} d$. 

Show that $\iota^*(A_{n+1} / \mathcal{I})$ is finitely generated over $A_n$.


\newpage

\textbf{Exercise 4.5}

Let $F: X \to X \times X$ be the polynomial map defined by $F(X) = (X, X)$. 

This is a generalized embedding in the sense of §3. 

Let $M$ and $N$ be left modules over $A_n$. 

Then $M \widehat{\otimes} N$ is a left module over $A_{2n} = A_n \widehat{\otimes} A_n$. 

Show that $F^*(M \widehat{\otimes} N) \cong M \otimes_{K[X]} N$. 
The latter was defined in Exercise 14.4.4.

\newpage

\textbf{Exercise 4.6}

Show that if $\iota: X \to X \times Y$ is the standard embedding, then
\[
\iota^*(A_{m+n}) \cong K[\partial_y] \widehat{\otimes} A_n
\]
where $K[\partial_y]$ is the left $A_m$-module $K[\partial_{y_1}, \ldots, \partial_{y_m}]$.






\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}


